Problem: Is ${947845}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {947845}= &&{9}\cdot100000+ \\&&{4}\cdot10000+ \\&&{7}\cdot1000+ \\&&{8}\cdot100+ \\&&{4}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {947845}= &&{9}(99999+1)+ \\&&{4}(9999+1)+ \\&&{7}(999+1)+ \\&&{8}(99+1)+ \\&&{4}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {947845}= &&\gray{9\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {9}+{4}+{7}+{8}+{4}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${947845}$ is divisible by $3$ if ${ 9}+{4}+{7}+{8}+{4}+{5}$ is divisible by $3$ Add the digits of ${947845}$ $ {9}+{4}+{7}+{8}+{4}+{5} = {37} $ If ${37}$ is divisible by $3$ , then ${947845}$ must also be divisible by $3$ Add the digits of ${37}$ $ {3}+{7} = \color{#9D38BD}{10} $ If $\color{#9D38BD}{10}$ is divisible by $3$ , then ${37}$ must also be divisible by $3$ $\color{#9D38BD}{10}$ is not divisible by $3$, therefore ${947845}$ must not be divisible by $3$.